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- Homework Statement
- A sample of potassium chromate, K2CrO4, is analyzed and found to contain exactly 0.67 grams of potassium. What is the mass of this sample of potassium chromate?

- Relevant Equations
- Formula mass = (2 x K) + (1 x Cr) + (4 x O)

1 mole of K2CrO4 has formula mass of ((2 x K) + (1 x Cr) + (4 x O)) g

=> ((2 x 39.0983) + (1 x 51.9961) + (4 x 15.9994)) g = 194.1903 g

we have 2 moles of K in 1 mole of K2CrO4 = (2x 39.0983) g = 78.1966 g

so for 1 mole of substance we have 78.1966 g

So how many moles of the substance do we have if our sample contains just 0.67 g of K

=> 1 mole x (0.67/78.1966) = 0.00856 moles

So the mass of K2CrO4 in 0.00856 moles is ( 0.00856 x194.1903) g

=> Mass of K2CrO4 of this sample is 1.66386 g

Now I was not sure of the number of significant figures required for the solution

since we were originally given the amount 0.67 g of K

I was unsure if this is to 2 or 3 significant figures so I gave the answer to both

I put down 1.66 g (to 3 sig fig) and 1.7 g (to 2 sig fig)

But both answers appear to be incorrect … can anyone help I am stuck ?

=> ((2 x 39.0983) + (1 x 51.9961) + (4 x 15.9994)) g = 194.1903 g

we have 2 moles of K in 1 mole of K2CrO4 = (2x 39.0983) g = 78.1966 g

so for 1 mole of substance we have 78.1966 g

So how many moles of the substance do we have if our sample contains just 0.67 g of K

=> 1 mole x (0.67/78.1966) = 0.00856 moles

So the mass of K2CrO4 in 0.00856 moles is ( 0.00856 x194.1903) g

=> Mass of K2CrO4 of this sample is 1.66386 g

Now I was not sure of the number of significant figures required for the solution

since we were originally given the amount 0.67 g of K

I was unsure if this is to 2 or 3 significant figures so I gave the answer to both

I put down 1.66 g (to 3 sig fig) and 1.7 g (to 2 sig fig)

But both answers appear to be incorrect … can anyone help I am stuck ?